Let A and B be two finite sets. Let f: A -->B be any map. Assume that |A|=|B|. (i) Show that if f is surjective, then f is injective. (ii) Show that if f is injective, then f is surjective. Solution i) Let f be surjective. Let f not be injective Let, a and b be two distinct elemetns in A so that f(a)=f(b) IN a map one element can map to only one element Hence, remaining |A|-2 elements other than a and b map to at most |A|-2=|B|-2 elements in B a and b map to f(a) Hence , f maps to at most |B|-2+1 =|B|-1 elements in B But f is sujrective to |f(A)|=|B| HEnce contradiction So, f is injective ii) Let, f be injective Let, |A|=|B|=n Let, elemetns of A as a1,..,an such ai is not equal to aj with i is not equal to j and elements of B as b1,b2,...bn so tha f(ai)=bi f is injective so f(ai) is not equal to f(aj) for i not equal to j Hence, b1,...bn are are mutually distinct Hence, f(A) has n=|B| elements Hence, f is surjective.